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3 years ago

Can someone help me solve for y?? Plsss

Mathematics

1 answer:

Kruka [31]3 years ago

8 0

Answer:

y>=-3/2x-5/2

Step-by-step explanation:

-1/2y<=-3/4x+5/4

y>=-3/2x-5/2

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Equation<br> Number of Solutions<br> 6(2.1 + 1) - 2 = 12.1 + 4<br> How many solutions does it have?

jek_recluse [69]

Infinite solutions!!!!

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3 years ago

0.002 is 1/10 of what

cestrela7 [59]

0.002 is the 1/10 of 0.02 number.

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3 years ago

- From a standing position, Meg jumps

Readme [11.4K]

Step-by-step explanation: Victor jumps farther than Meg.

If Victor jumps 9 feet and 2 inches and Meg jumps 7 feet and 4 inches, that means that Victor jumps 1 foot 10 inches farther than Meg.

Lets learn our conversations: 1 foot = 12 inches

If Victor jumps 9 feet, that means that 9 feet = 108 inches + 2 inches, which equals 110 total inches jumped.

If Meg jumps 7 feet, that means that 7 feet = 84 inches + 4 inches, which equals 88 total inches jumped.

110 - 88 = 22 inches.

So, Victor jumped 22 inches farther than Meg, or 1 foot and 10 inches farther than Meg.

8 0

4 years ago

A cake is removed from a 310°F oven and placed on a cooling rack in a 72°F room. After 30 minutes the cake's temperature is 220°

Fynjy0 [20]

Answer:

The time is 135 min.

Step-by-step explanation:

For this situation we are going to use Newton's Law of Cooling.

Newton’s Law of Cooling states that the rate of temperature of the body is proportional to the difference between the temperature of the body and that of the surrounding medium and is given by

$T(t)=C+(T_0-C)e^{kt}$

where,

C = surrounding temp

$T(t)$ = temp at any given time

t = time

$T_0$ = initial temp of the heated object

k = constant

From the information given we know that:

Initial temp of the cake is 310 °F.
The surrounding temp is 72 °F.
After 30 minutes the cake's temperature is 220 °F.

We want to find the time, in minutes, since the cake's removal from the oven, at which its temperature will be 100°F.

To do this, first, we need to find the value of k.

Using the information given,

$220=72+(310-72)e^{k\cdot 30}\\\\72+238e^{k30}=220\\\\238e^{k30}=148\\\\e^{k30}=\frac{74}{119}\\\\\ln \left(e^{k\cdot \:30}\right)=\ln \left(\frac{74}{119}\right)\\\\k\cdot \:30=\ln \left(\frac{74}{119}\right)\\\\k=\frac{\ln \left(\frac{74}{119}\right)}{30}$

$T(t)=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}$

Next, we find the time at which the cake's temperature will be 100°F.

$100=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}\\72+238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=100\\238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=28\\e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=\frac{2}{17}\\\ln \left(e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}\right)=\ln \left(\frac{2}{17}\right)\\\frac{\ln \left(\frac{74}{119}\right)}{30}t=\ln \left(\frac{2}{17}\right)\\t=\frac{30\ln \left(\frac{2}{17}\right)}{\ln \left(\frac{74}{119}\right)}\approx 135.1$